An explicit isomorphism between the 1st Cech cohomology and the 1st hypercohomology

Question

Let $\mathbf{X}$ be a Grothendieck topos and let $A$ be an abelian group in $\mathbf{X}$. Verdier's Theorem allows one to describe $\mathrm{H}^n(\mathbf{X},A)$ in terms of hypercoverings, namely, as the colimit of $\mathrm{H}^n(U_\bullet,A)$ where $U_\bullet$ ranges over the category of hypercoverings (of the terminal object of $\mathbf{X}$).

One can further form the Cech cohomology $\check{\mathrm{H}}^n(\mathbf{X},A)$, which is the colomit of $\mathrm{H}^n(U_\bullet,A)$ as $U_\bullet$ ranges over hypercoverings with $U_\bullet=\mathrm{cosk}_0(U_\bullet)$. Let me call them Cech hypercoverings. (These are the hypercoverings with $U_n=U_0^{\times (n+1)}$ and the standard simplicial structure.)

A well-known spectral sequence relates the derived-functor cohomology with the Cech cohomology, implying in particular that $\mathrm{H}^1(\mathbf{X},A)$ is canonically isomorphic to $\check{\mathrm{H}}^1(\mathbf{X},A)$. I am not absolutely certain, but it seems correct that under the description provided by Verdier's Theorem, the isomorphism $\check{\mathrm{H}}^1(\mathbf{X},A)\to \mathrm{H}^1(\mathbf{X},A)$ is the obvious one, namely, if a Cech cohomology class is represented by a $1$-cocycle in $Z^1(U_\bullet,A)$, then its image in $\mathrm{H}^1(\mathbf{X},A)$ is the cohomology class represented by that $1$-cocycle.

The latter statement means that for a hypercovering $U_\bullet$ (not necessarily Cech) and any $1$-cocycle $\alpha\in Z^1(U_\bullet,A)$, one can find a Cech hypercovering $U'_\bullet$ and $\alpha'\in Z^1(U'_\bullet,A)$ representing the same cohomology class. I am looking for a way to construct these $U'$ and $\alpha'$ directly. The main problem is that $U_\bullet$ cannot be refined to a Cech hypercovering in general.

An alternative approach to the problem (which also applies to non-abelian $A$) is via the correspondence with $A$-torsors. It is a standard fact that $\check{H}^1(\mathbf{X},A)$ is in 1-1 correspondence with isomorphism classes of $A$-torsors. Explicitly, if $\alpha\in Z^1(U_\bullet,A)$ with $U_\bullet$ being a Cech hypercovering, then the $A$-torsor $P$ corresponding to $\alpha$ can be described by $$ P(V)=\{a\in A(U_0\times V)~:~ \alpha_V \cdot d_0^1a =d_1^1 a\} $$ where $d^1_i:A(U_0\times V)\to A(U_1\times V)=A(U_0\times U_0\times V)$ is induced by $d^1_i:U_1\to U_0$. Suppose now that $U_\bullet$ is an arbitrary hypercovering (not necessarily Cech). The question will be resolved if one can show that $P$ constructed above is still an $A$-torsor. The difficult thing to check is that $P\neq \emptyset$. (However, when $U_\bullet$ is Cech, one can check that $\alpha\in P(U_0)$.)

Answer 1

Some hints in the literature led me to an answer, which I find a bit surprising: One can take $U'_\bullet=\mathrm{cosk}_0(U_\bullet)$ and the $1$-cocycle $\alpha\in Z^1(U_\bullet,A)\subseteq A(U_1)$ descends uniquely to a $1$-cocycle in $\alpha'\in Z^1(\mathrm{cosk}_0(U_\bullet),A)\subseteq A(U_0\times U_0)$ along $(d_0,d_1):U_1\to U_0\times U_0$. In other words:

Proposition: For any hypercovering $U_\bullet$, the canonical map $Z^1(\mathrm{cosk}_0(U_\bullet),A)\to Z^1(U_\bullet,A)$ is an isomorphism. Consequently, the map $\mathrm{H}^1(\mathrm{cosk}_0(U_\bullet),A)\to \mathrm{H}^1(U_\bullet,A)$ is an isomorphism.

This is a nontrivial statement so let me sketch the ad-hoc proof I have.

Step 1: We may assume $U_\bullet=\mathrm{cosk}_1(U_\bullet)$.

Indeed, since $U_\bullet$ is a hypercovering, the map $U_2\to \mathrm{cosk}_1(U_\bullet)_2$ is a covering, and this easily implies that $Z^1(\mathrm{cosk}_1(U_\bullet),A)\to Z^1(U_\bullet,A)$ is an isomorphism, so replace $U_\bullet$ with its $1$-coskeleton.

One consequence of this assumption is that $U_2=\underline{\mathrm{Hom}}_{\mathrm{Simp}}(\partial \Delta^2, U_\bullet)$, where $\partial \Delta^2$ is the boundary of the $2$-simplex, realized as a constant simplicial object (i.e. sheaf) in $\mathbf{X}$, and $\underline{\mathrm{Hom}}$ denote internal $\mathrm{Hom}$ in $\mathbf{X}$.

Step 2: Let $\alpha\in Z^1(U_\bullet,A)$. We claim that $a\in A(U_1)$ descends along $(d_0,d_1):U_1\to U_0\times U_0$ to some $\alpha'\in G(U_0\times U_0)$.

Let $V=U_1\times_{U_0\times U_0}U_1$ and let $\pi_1,\pi_2:V\to U_1$ denote the first and second projections. Let $S$ denote the simplicial object of $\mathbf{X}$ obtained by gluing two copies of $\Delta^1$ along their vertices. Then $V=\underline{\mathrm{Hom}}_{\mathrm{Simp}}(S, U_\bullet)$. There is a simplicial map $\partial \Delta^2\to S$ which degenerate the edge $\{1,2\}$ into a vertex. This gives rise to a map $V\to U_2$, which in turn gives rise a map $A(U_2)\to A(V)$. Applying this map to the cocycle equation $d_0^2\alpha-d_1^2\alpha+d^2_2\alpha=0$ in $A(U_2)$ gives $0-\pi_2^*\alpha+\pi_1^*\alpha=0$ in $A(V)$, which means that $\alpha$ descends to $\alpha'\in G(U_0\times U_0)$.

Step 3: We finally claim that $\alpha'$ lies in $Z^1(\mathrm{cosk}_0(U_\bullet),A)$, which proves the surjectivity of $Z^1(\mathrm{cosk}_0(U_\bullet),A)\to Z^1(U_\bullet,A)$. The injectivity follows easily from the fact that $(d_0,d_1):U_1\to\mathrm{cosk}_0(U_\bullet)_1=U_0\times U_0$ is a covering.

To show the claim, it is enough to show that the canonical map $U_2\to \mathrm{cosk}_0(U_\bullet)_2=U_0\times U_0\times U_0$ is a covering. Indeed, if this holds, then the fact that the $1$-cocycle equation holds for $\alpha$ in $A(U_2)$ implies that it holds for $\alpha'$ in $A(U_0\times U_0\times U_0)$. Proving that $U_2\to U_0\times U_0\times U_0$ is a covering amounts to showing that any $3$ "vertices" in $U_0$ can be joined by a "triangle" in $U_2$ locally. But this follows from $U_2=\underline{\mathrm{Hom}}_{\mathrm{Simp}}(\partial \Delta^2, U_\bullet)$ and the fact that $(d_0,d_1):U_1\to U_0\times U_0$ is a covering.

I would value references for this proposition in the literature, if you know them.